For many purposes we can treat butane (C4H10) as an ideal gas at temperatures above

A. Yes

B. –176 °C

Explanation:

A. Yes

B. Determination of the new temperature of the gas.

Let the initial pressure be P

From the question given above, the following data were obtained.

Initial pressure (P1) = P

Initial temperature (T1) = 19 °C

Final pressure (P2) = ⅓ P1 = ⅓P = P/3

Final temperature (T2) =?

Next, we shall convert 19 °C to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T1) = 19 °C

Initial temperature (T1) = 19 °C + 273

Initial temperature (T1) = 292 K

Since the volume is constant, we can obtain the new temperature of the gas as illustrated below:

Initial pressure (P1) = P

Initial temperature (T1) = 292 K

Final pressure (P2) = P/3

Final temperature (T2) =?

P1/T1 = P2/T2

P/292 = P/3 /T2

P/292 = P/3T2

Cross multiply

P × 3T2 = 292 × P

Divide both side by P

3T2 = (292 × P)/P

3T2 = 292

Divide both side by 3

T2 = 292/3

T2 = 97.33 ≈ 97 K

Finally, we shall convert 97 K to celcius temperature. This can be obtained as follow:

T(°C) = T(K) – 273

T(K) = 97

T(°C) = 97 – 273

T(°C) = –176 °C

Thus, the new temperature of the gas is –176 °C.

c.

explanation:

i think that the answer is c